PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Direct Current and Transient Analysis 179

ANALYTICAL Solution

Part a

The loop differential equation, as well as its current solution i(t), is given as follows:

L

di t

dt

Ri t V

()


() 0

then

it

V

R

V

R

()

^00 et Rt for 0 for 1, 2, and 5

where τ = L/R.

MATLAB Solution

% Script file: RL

% parts ( b and c )

itR _ 1 = dsolve(‘Dy+y =10’,’y(0) = 0’,’t’);

itR _ 2 = dsolve(‘Dy+2*y=10’,’y(0)=0’,’t’);

itR _ 5 = dsolve(‘Dy+5*y=10’,’y(0)=0’,’t’);

disp(‘***************************’);

disp(‘The solutions i(t) for R=1,2 and 5 (in Ohms) are:’);

disp(‘***************************’);

disp(‘***************************’);

itR _ 1

disp(‘***************************’);

itR _ 2

disp(‘***************************’);

itR _ 5

disp(‘***************************’);

disp(‘***************************’);

ezplot(itR _ 5,[0 3]);

hold on;

grid on; ezplot(itR _ 2,[0 3]);

hold on;

ezplot(itR _ 1,[0 3]);

hold on;

title(‘i(t) vs t for L=1 & R=1,2 and 5’)

axis([0 3 0 11]);

xlabel(‘time t’);ylabel(‘i(t)’)

FIGURE 2.81

Network of Example 2.23.

V 0 = 10 V R

Switch closes at t = 0 L = 1 H

i(t)